# Gluing submanifolds along their common boundary

Let $M$ be a smooth $m$-manifold and $N_1$, $N_2$ smooth embedded $k$-submanifolds such that

• $N_1\cap N_2=\partial N_1=\partial N_2$,
• for each $x\in N_1\cap N_2$, $T_x N_1=T_x N_2$, and
• for $x\in N_1\cap N_2$, each vector in $T_x N_1$ is inwards-pointing wrt. $N_1$ iff it is outwards pointing wrt. $N_2$ and vice versa.

Is it true that $N_1\cup N_2$ is a smooth embedded $k$-submanifold with $N_1\cap N_2$ in its interior?

Edit: It follows from mollyerin's answer that $N$ is not necessarily the image of a $C^\infty$ embedding. But I have further bounty questions:

• Is it at least the image of a $C^1$ embedding?
• Are there some further simple conditions that I should include so that $N$ is a smooth ($C^\infty$) embedded submanifold?
• Can I "change" $N$ in some arbitrary small neighborhood of $N_1\cap N_2$ to make the resulting manifold smooth?
• Could you please comment on the downvote? Jan 23, 2015 at 21:00
• I didn't downvote, but there is a faction on this site who think it's not okay to ask a question unless you show some work that you've done. Jan 23, 2015 at 21:05
• Ok, right, thanks. I will work on the quesiton. Jan 23, 2015 at 21:05
• Consider something like this: N1 be the upper branch of $y=x^3$, and N2 be the lower branch of $y=-x^3$. Jan 23, 2015 at 23:48
• @XipanXiao Right, thanks. Of course I should add that invards pointing vectors in $T_x N_1$ wrt. $N_1$ are outwards pointing wrt. $N_2$. Now is it sufficient? Jan 24, 2015 at 0:03

This is false; you'll need assumptions (I'm not sure what good ones are) that ensure that the manifolds glue together "smoothly". As a counterexample, let $f_1: (-\infty, 0] \to \mathbb{R}^2$ be the map $f_1(x) = -x^2$, and let $f_2 : [0, \infty) \to \mathbb{R}^2$ be given by $f_2(x) = x^2$. We of course let $N_1$ be the image of $f_1$ and $N_2$ the image of $f_2$. Then $N_1$ and $N_2$ satisfy your conditions. However, $N = N_1 \cup N_2$ is not a smooth embedded manifold.
[Proof: If it were, there would be a smooth embedding $g : \mathbb{R} \to \mathbb{R}^2$ whose image is $N$. Say $g = (g_1, g_2)$. The map $g_1$ satisfies $g_1'(t) \neq 0$ for all $t$, and so (since it is also smooth), it is a diffeomorphism $\mathbb{R} \to \mathbb{R}$ (by the inverse function theorem). Therefore the map $g \circ g_1^{-1}$ is a smooth embedding $\mathbb{R} \to \mathbb{R}^2$ whose image is $N$, and this map has the form $x \mapsto (x, g_2 \circ g_1^{-1}(x)$), so evidently $g_2 \circ g_1(x) = sign(x) x^2$. But this map is not smooth. ]
• Thanks. Can I at least say that $N$ is a $C^1$-submanifold, or something like that? Jan 24, 2015 at 11:12
• Seems like $N$ should be the image of a $C^1$ embedding of a smooth manifold, but the proof isn't immediately obvious to me (I get confused about these things); you might try unaccepting my answer and editing the original question to see if anyone has ideas or a reference. (I will also try thinking about it.) Jan 24, 2015 at 11:30